How do you find all solutions to #x^3-(1-i)=0#?

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Answer 1 · 2016-12-31T12:21:43

Roots are:

#2^(1/6) exp(- (i pi) /12)#

#2^(1/6) exp( (7i pi) /12)#

#2^(1/6) exp( (5 i pi)/4 )#

you are looking for:

#x = (1-i)^(1/3)#

as a rule, this complex number will have 3 cube roots, same absolute value but spaced equally about the Argand diagram.

using #cis (theta) equiv e^(i theta)# in the notation, we can express this in polar coordinates as follows:

#1-i = \sqrt 2 \ cis(- pi /4 color(red)(+ 2n pi))# where #n in Z#

So

#(1-i)^(1/3) = (\sqrt 2 \ cis(- pi /4 + 2n pi))^(1/3)#

# = 2^(1/6) \ cis(- pi /12 + (2n pi)/3 )#

working through values of n

#n = 0 to 2^(1/6) cis(- pi /12) = 2^(1/6) exp(- (i pi) /12)#

#n = 1 to 2^(1/6) cis( (7pi) /12) = 2^(1/6) exp( (7i pi) /12)#

#n = 2 to 2^(1/6) cis( (5pi)/4 ) = 2^(1/6) exp( (5 i pi)/4 )#

and then the roots start repeating as follows
#n = 3 to 2^(1/6) cis(- pi /12 + 2pi ) = 2^(1/6) cis(- pi /12 )#